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The mole fraction of solute in a 3.87 m aqueous solution is 0.0697
<h3>
calculation</h3>
molality = moles of the solute/Kg of the solvent
3.87 m dissolve in 1 Kg of water= 1000g
find the moles of water= mass/molar mass
that is 1000 g/ 18 g/mol= 55.56 moles
mole of solute = 3.87 moles
mole fraction is = moles of solute/moles of solvent
that is 3.87/ 55.56 = 0.0697
Answer:
44,55 can be produced.
Explanation:
First, we balanced the equation
1Cu + 2AgNO3 → 1Cu(NO3)2 + Ag
Then, we find the moles of each reagent


Now, we find the limiting reagent from the quantities of product that can be formed from each reagent


1,086 moles of AgNO3 is necessary for each mole of Cu since we have 0.413 moles of Ag(NO3), the nitrate is the limiting reagent
the value of the limiting reagent determines the amount of product that is generated
∴ 0,413 mol of Ag can be produced
Ag =
= 44,55g
Ag≈ 44,6g